Analog circuits Test 2

If the early effect is considered, the low frequency response of the C.E. stage becomes g_m*(R_l || ro). Thereby, miller approximation shows that the capacitance between the base and the collector, looking from the output side, will be increased by a factor of 1 + 1/g_m*(R_l || ro). Hence the correct option is 1 + 1/ g_m*(R_l || ro).

We have a capacitor from the collector to substrate, C_cs, which comes in parallel to the miller approximation of the capacitance from base to collector. The miller approximation defines the latter as C_µ*(1 + 1/g_m*R₂). Since capacitors gets added, when in parallel, the correct option is C_cs + C_µ*(1+ 1/g_m*R₂).

All the parasitic capacitors of a B.J.T. affect the C.B. stage. None of the parasitic capacitors gets deactivated and they end up behaving as a pole during the frequency response of the C.B. stage.

The transit frequency is directly proportional to the transconductance of the B.J.T. Hence, the correct option is increases. Since it hasn’t been mentioned that whether the transconductance has been doubled or not, we cannot conclude the option “doubles” as an answer.

The constant K = V₂/V₁, which is the internal voltage gain of the network.

Thus resistance R_M = R/1-K

R_N = R/1-K^-1.

Apply millers theorem to resistance between input and output.
At input, R_M = 100k/1-K = R_I
Output, R_N = 100k/1-K^-1 ≈ 100k
Internal voltage gain , K = -h_feR_L’/h_ie
K = – 50xR_c||100k/1k = – 50x4x100/104 = – 192
R_I = 100k/1+192 = 0.51kΩ
R_I’ = R_I||h_ie = 0.51k||1k = 0.51×1/1.51 = 0.337kΩ
Net voltage gain = K.R_I’/R_S+R_I’ = – 192 x 0.337/2k + 0.337k = -27.68.

A_V = -120

f_L = 1/2πC_C(R_C + R_L) = 1/2π x 0.001 = 1000/2π = 159.15Hz

A_V’ = 120/8.02 ≈ 15.

R_C = 2kΩ, R_L = 5kΩ, C_C = 1μF, C_B = 10μF, C_E = 20μF, R_S = 2 kΩ
h_ie = 1kΩ, I_C = 2mA
f_L1 = 1/2πC_C(R_C+R_L) = 22.73 Hz
f_L2 = g_m/2πC_E = I_C/2πC_EV_T = 612 Hz
Since f_L2 > 4f_L1, hence f_L2 is the correct answer.

Total phase shift = 180°+ tan^-1(f_L/f)

At 3dB frequency f_L/f = 1

Total phase shift = 180° + 45° = 225°.

Given R1 > R2
R/1-K > R/1-K-1, and so 1-K-1 > 1-K
Thus K2>1, K>1, K<-1 (correct)
Thus, C1 = C(1-K) and C2 = C(1-K-1)
Hence C1>C2.

C1 = C(1 - K), C2 = C(1 - K^-1)
C1 = 2pF
C2 = 4pF
C1/C2 = 1/2 = 1-K/1-K-1
K = -2
C1 = C(1 + 2) = 3C
C = C1/3 = 2/3pF = 0.67 pF.

A_VM = 150
A_VL = 100
f = 50Hz
1 + f²/2500 = 1.5²
f² = 2500*1.25 = 3125
f = 55.90 Hz.

Net load = 10k||10k = 5kΩ = R_L’

A_VM = -h_feR_L’/h_ie = -50 × 5/1 = -250

f_L = 1/2πC_C(R_C+R_L) = 15.9 Hz

A_VL = 133.

180° + tan^-1(f_L/f) = 260°

f_L/f = tan(80) = 5.67

A = 20.3 + 5.672 = 26.05.

Concept:

\({Z_1} = {R_1}\)

\({Z_2} = {R_2}||C\)

\({Z_2} = \frac{{{R_2}\left( {\frac{1}{{j\omega c}}} \right)}}{{{R_2} + \frac{1}{{j\omega c}}}}\;\)

\({Z_2} = \frac{{{R_2}}}{{1 + j\omega {R_2}C}}\)

For the inverting amplifier:

\(\frac{{{V_o}}}{{{V_i}}} = - \frac{{{Z_2}}}{{{Z_1}}}\)

\(\frac{{{V_o}}}{{{V_i}}} = \frac{{ - \frac{{{R_2}}}{{1\_j\omega {R_2}C}}}}{{{R_1}}} = \frac{{\frac{{{R_2}}}{{{R_1}}}}}{{1 + j\omega {R_2}C}}\)

The magnitude of the voltage gain will be:

\(\left| {\frac{{{V_o}}}{{{V_i}}}} \right| = \frac{{\frac{{{R_2}}}{{{R_1}}}}}{{\sqrt {1 + {{\left( {\omega {R_2}C} \right)}^2}} }}\)

At ω = 0:

\(\left| {\frac{{{V_o}}}{{{V_i}}}} \right| = \frac{{{R_2}}}{{{R_1}}}\)

Let it be A_max

\(\therefore \frac{{{v_o}}}{{{V_i}}} = \frac{{{A_{msx}}}}{{1 + j\omega {R_2}C}}\)

\(\left| {\frac{{{V_o}}}{{{V_i}}}} \right| = \frac{{{A_{max}}}}{{\sqrt {1 + {{\left( {\omega {R_2}C} \right)}^2}} }}\)

At ω = ω_c (Cut-off frequency)

\(\left| {\frac{{{V_o}}}{{{V_i}}}} \right| = \frac{{Amax}}{{\sqrt 2 }}\)

\(\frac{{Amax}}{{\sqrt 2 }} = \frac{{Amax}}{{\sqrt {1 + {{\left( {{\omega _c}{R_2}C} \right)}^2}} }}\)

2 = 1 + (ω_cR₂C)²

1 = ω_cR_2­C

\({\omega _c} = \frac{1}{{{R_2}C}}\)

The frequency response is plotted as:

Calculation:

\({f_c} = \frac{1}{{2\pi {R_2}C}}\)

\({R_2} = \frac{1}{{2\pi {f_c}C}}\)

Putting on the respective values, we get: 

\({R_2} = \frac{1}{{2\pi \; \times \;5\; \times\; {{10}^3}\; \times \;10\; \times\; {{10}^{ - 9}}}}\)

R₂ = 3.10 kΩ

Note: It is important to understand the formula for cut-off frequency and remember it for once. It saves a lot of time without the analysis to conclude that:

\({\omega _c} = \frac{1}{{{R_2}C}}\)

M₂ behaves as a Source follower i.e. it is being used in a C.D. configuration. M₁ provides a output resistance of r_o1 but M₂ provides an impedance of (1/g_m2||r_o2). Thus, from the voltage gain of a follower stage, we conclude the overall voltage gain is r_o1 / {(1/g_m1 || r_o2) + r_o1}.

The output impedance of the follower is a function of the transconductance. But it’s also a function of the channel length modulation. But the resistance offered due to channel length modulation is much greater than the inverse of the transconductance. It can be concluded that the output of the follower depends on the transconductance and it’s low.

The voltage gain is given by R_L/(1/g_m + R_L) where R_L is the 50 Ω load. Now, we see that if the voltage gain is .25, gm is 1/175. Now, gm is  where (W/L) is the aspect ratio, Id is the drain current. We have all the values and the aspect ratio becomes 32.6.

The source of M₁ and M₂ are connected at node X. The impedance looking into the source of a MOSFET is (1/gm || ro). If we look at node X, M₁ and M₂ are connected from Node X to ground. Hence, they are parallel to each other and the overall impedance becomes (1/g_m2 || r_o2 || 1/g_m1 || r_o1).

The input to the follower stage is applied to the gate of the MOSFET. The gate of the MOSFET is made of SiO₂, which is a dielectric and provides very high impedance to the input.

The C.D. configuration offers high input impedance to the input signal and low output impedance while sensing the output signal. Thus, it is used as a buffer stage.

Since the transconductance is .02, it’s inverse is 50. From the general expression of voltage gain of a source follower, we conclude that the load should be 50 Ω since the inverse of the transconductance is the impedance looking into the source of the MOSFET. For maximum power transfer, we have to match both the impedance.

The gain of the follower stage is given by R_s/(1/g_m + R_s). Hence, we readily conclude that if the transconductance (g_m) increases, the gain will increase.

Formula used:

V_DC = I_DCR_L

Ripple voltage is calculated as:

\({V_r} = \frac{{{I_{DC}}}}{{{f_o}C}}\) 

Ripple factor, r is given by:

\(r = \frac{1}{{2\sqrt 3 {f_o}C{R_L}}}\) 

Also, the angle of conduction is:

\(= \sqrt {\frac{{2{V_r}}}{{{V_m}}}} \) 

\({V_m} = {V_{DC}} + \frac{{{V_r}}}{2}\) 

Calculation:

Given: V_DC = 40 V, R_L = 2 kΩ, f_o = 60 Hz, and C = 680 μf

\({I_{DC}} = \frac{{{V_{DC}}}}{{{R_L}}} = \frac{{40}}{2}mA\) 

I_DC = 20 mA

Now, the ripple voltage will be”

\({V_r} = \frac{{{I_{DC}}}}{{{f_o}C}} = \frac{{20 \times {{10}^{ - 3}} \times {{10}^6}}}{{60 \times 680}}\) 

V_r = 0.49 V

\({V_m} = {V_{DC}} + \frac{{{V_r}}}{2} = 40 + \frac{{0.49}}{2}\) 

V_m = 40.24 V

The angle of conduction will now be:

\(\sqrt {\frac{{2{V_r}}}{{{V_m}}}} = \sqrt {\frac{{2 \times 0.49}}{{40.24}}} \) 

= 0.156 radian

Duty cycle \(= \frac{{angle\;of\;conduction}}{{2\pi }} \times 100\%\) 

\(= \frac{{0.156}}{{2\pi }} \times 100\%\) 

= 2.48%

Options 1) & 2) are correct

The Ripple factor for capacitor filler will be:

\(r = \frac{1}{{2\sqrt 3 {f_o}C{R_L}}}\) 

\(r \propto \frac{1}{{{R_L}}}\) 

For large R_L, the ripple factor will be smaller.

Also, since the Peak to peak ripple voltage is:

\({V_r} = \frac{{{I_{DC}}}}{{{f_o}C}}\) 

\({V_r} \propto \frac{1}{{{f_o}}}\) 

Peak to peak ripple voltage decreases with frequency.

Options 1, 2 & 3 are correct.